输入一个数字，问从1开始到这个数字总共写了多少个数字

本文链接：https://blog.youkuaiyun.com/BHliuhan/article/details/81219282

这道题目要求计算从1开始到给定图书馆书籍数量的数字总个数。例如，如果有13本书，需要从1到13进行编号，总共会写17个数字。给出一个整数n，输出从1到n编号所需的总数字个数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

B. Vanya and Books
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.

Vanya wants to know how many digits he will have to write down as he labels the books.

Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.

Output
Print the number of digits needed to number all the books.

Examples
inputCopy
13
outputCopy
17
inputCopy
4
outputCopy
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.

Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.

THINK:
举个例子，1236，算的时候是（1236-1000+1）*4+（1000之前的所有数字，不包含1000）
100-999之间是900个数，900*3；10-99之间是90个数，90乘2，1-9之间是9个数，这些所有的加起来；最后用代码实现出来
重要的是自己的思路，自己想的太复杂，相同的三位数就不要单独分析了，就把他们都归类；
另外.pow函数是double 类型；不能转化成long long int 类型

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
long long int po(long long int n,long long int m)
{
    long long int s=1;
    for(int i=1;i<=m;i++)
    {
        s*=n;
    }
    return s;
}
long long int f(long long int n,long long int s)
{
    long long int sum=0;
    if(s==1)sum=n;
    else
    {
        for(long long int i=1;i<=s-1;i++)
        {
            sum+=9*po(10,i-1)*i;
        }
        sum+=(n-po(10,s-1)+1)*s;
    }
    return sum;
}
int main()
{
    long long int n,t;
    scanf("%lld",&n);
    t=n;
    long long int s=0,sum=0;
    while(t!=0)
    {
        t=t/10;
        s++;
    }
    printf("%lld\n",f(n,s));
    return 0;
}